Question:
If the tangent to the curve, $y=f(x)=x \log _{e} x$, $(x>0)$ at a point $(c, f(c))$ is parallel to the line - segement joining the points $(1,0)$ and $(e, e)$, then $\mathrm{c}$ is equal to :
Correct Option: , 3
Solution:
$f(x)=x \log _{e} x$
$\left.f^{\prime}(\mathrm{X})\right|_{(\mathrm{c}, f(\mathrm{c}))}=\frac{\mathrm{e}-0}{\mathrm{e}-1}$
$f^{\prime}(\mathrm{x})=1+\log _{\mathrm{e}} \mathrm{x}$
$\left.f^{\prime}(\mathrm{x})\right|_{(\mathrm{c}, f(\mathrm{c}))}=1+\log _{\mathrm{e}} \mathrm{c}=\frac{\mathrm{e}}{\mathrm{e}-1}$
$\log _{e} c=\frac{e-(e-1)}{e-1}=\frac{1}{e-1} \Rightarrow c=e^{\frac{1}{e-1}}$
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